Question

Find the non-negative root of ${\left(\frac{3x-1}{2x+3}\right)}^2-5\left(\frac{3x-1}{2x+3}\right)+4=0$

• A. 2
• B. 5
• C. 4
• D. 3

Answer 1

The correct option is C

4

Let $\frac{3x-1}{2x+3}=y$

The equation reduces to:

$y^2-5y+4=0$

$\Rightarrow (y-1)(y-4)=0$
$\Rightarrow y=1,4$

Now, $\frac{3x-1}{2x+3}=1$
$\Rightarrow 3x-1=2x+3$
$\Rightarrow x=4$

and $\frac{3x-1}{2x+3}=4$
$\Rightarrow 3x-1=8x+12$
$\Rightarrow x=\frac{-13}{5}$